Assume F is a field of characteristic different than 2. Let a, b be invertible elements in F, and let A(a,b) be the generalised quaternions. Using the Artin–Wedderburn theorem, there is a representation of A(a,b) over F. I found a representation as Q8 but it's not over F. So, how to find a representation as matrices over F?
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3$\begingroup$ Have you looked at en.wikipedia.org/wiki/Quaternion_algebra ? $\endgroup$– S. Carnahan ♦Commented Jun 1, 2013 at 22:38
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2 Answers
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You could take the regular representation (left multiplication on $A$). So if $x^2 = a, y^2 = b$ then taking a basis $\{1,x,y,xy\}$ of $A$, $x$ would be represented by the matrix $$\left( \begin{array}{cccc} 0 & a & 0 & 0 \cr 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & a \cr 0 & 0 & 1 & 0 \end{array} \right), $$ etc.
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$\begingroup$ But Artin–Wedderburn theorem claim that there is 2*2 matrix over F exists. $\endgroup$– AndrewCommented Jun 1, 2013 at 12:52
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2$\begingroup$ The Artin-Wedderburn theorem doesn't claim there is a representation as 2 x 2 matrices. Take Hamilton's quaternions $A(-1,-1)$ with $F = {\mathbf R}$. The usual representation with matrices uses 4 x 4 real matrices or 2 x 2 complex matrices. $\endgroup$– KConradCommented Jun 1, 2013 at 12:57
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$\begingroup$ ok, how to represent generalised quaternions A(a,b) using 2*2 matrices? I need this, cuz I want to prove isomorphism between generalised quaternions over F and matrix 2*2 over F. Thanks! $\endgroup$– AndrewCommented Jun 1, 2013 at 13:04
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$\begingroup$ can we choose subfield closed so q^2=-a and p^2=-b hence we have representation i=((q,0),(0,-q)) and j=((0,p),(-p,0))? $\endgroup$– AndrewCommented Jun 1, 2013 at 13:12
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2$\begingroup$ An isomorphism between the generalized quaternion algebra $(a,b)_F$ and $M_2(F)$ only exists if the the form $x^2-ay^2-bz^2+abt^2$ has a nontrivial zero. $\endgroup$– NameCommented Jun 1, 2013 at 14:00
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You need a (necessarily non-central) sub-field $K$ of your quaternions $Q$, of degree $2$ over the center $k$ of the quaternions. Then there is a representation of $Q$ as $2$-by-$2$ matrices over $K$, by choosing any $K$-basis of $Q$...
answered Mar 3, 2015 at 20:08